To find the gradient of the scalar function \( f(x, y, z) = e^{xy} - x \cos(yz^2) \), we need to compute the partial derivatives of \( f \) with respect to each variable \( x \), \( y \), and \( z \).

The gradient of \( f \), denoted as \( \nabla f \), is a vector of partial derivatives:

\[
\nabla f = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right)
\]

### Step 1: Partial derivative with respect to \( x \)
\[
\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( e^{xy} - x \cos(yz^2) \right)
\]
- The derivative of \( e^{xy} \) with respect to \( x \) is \( y e^{xy} \).
- The derivative of \( -x \cos(yz^2) \) with respect to \( x \) is \( -\cos(yz^2) \).

So,
\[
\frac{\partial f}{\partial x} = y e^{xy} - \cos(yz^2)
\]

### Step 2: Partial derivative with respect to \( y \)
\[
\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( e^{xy} - x \cos(yz^2) \right)
\]
- The derivative of \( e^{xy} \) with respect to \( y \) is \( x e^{xy} \).
- The derivative of \( -x \cos(yz^2) \) with respect to \( y \) is \( x z^2 \sin(yz^2) \), using the chain rule.

So,
\[
\frac{\partial f}{\partial y} = x e^{xy} + x z^2 \sin(yz^2)
\]

### Step 3: Partial derivative with respect to \( z \)
\[
\frac{\partial f}{\partial z} = \frac{\partial}{\partial z} \left( e^{xy} - x \cos(yz^2) \right)
\]
- The derivative of \( e^{xy} \) with respect to \( z \) is \( 0 \) because there is no \( z \) in this term.
- The derivative of \( -x \cos(yz^2) \) with respect to \( z \) is \( 2xyz \sin(yz^2) \), using the chain rule.

So,
\[
\frac{\partial f}{\partial z} = 2xyz \sin(yz^2)
\]

### Final Answer:
The gradient of \( f \) is:

\[
\nabla f = \left( y e^{xy} - \cos(yz^2),\ x e^{xy} + x z^2 \sin(yz^2),\ 2xyz \sin(yz^2) \right)
\]

Let me know if you need further details or have any questions!

Here are 5 related questions:
1. How do we apply the chain rule when differentiating composite functions like \( \cos(yz^2) \)?
2. How would the gradient change if \( f(x, y, z) \) had a different form, such as \( e^{x^2y} \)?
3. What does the gradient of a scalar function represent geometrically?
4. How can we find the directional derivative of \( f \) in the direction of a given vector?
5. How would the gradient be different if we changed the trigonometric term in \( f \) to \( \sin(yz) \)?

**Tip:** The gradient vector points in the direction of the steepest increase of a scalar function.

Learn how to compute the gradient of the function f = e^xy - xcos(yz^2) by finding its partial derivatives with respect to x, y, and z. This step-by-step solution covers key concepts in multivariable calculus such as the gradient, chain rule, and product rule, making it suitable for college-level calculus students.

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